What is (\partial)/(\partial x)(tan(5+2x^2y^3z^2)) ?

The answer to (\partial)/(\partial x)(tan(5+2x^2y^3z^2)) is sec^2(5+2x^2y^3z^2)*4y^3z^2x

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(\partial)/(\partial x)(tan(5+2x^2y^3z^2))

\frac{\partial}{\partial x} (tan (5 + 2 x^{2} y^{3} z^{2}))

sec^{2} (5 + 2 x^{2} y^{3} z^{2}) \cdot 4 y^{3} z^{2} x

Solution steps

\frac{\partial}{\partial x} (tan (5 + 2 x^{2} y^{3} z^{2}))

Treat

y, z

as constants

Apply the chain rule:

sec^{2} (5 + 2 x^{2} y^{3} z^{2}) \frac{\partial}{\partial x} (5 + 2 x^{2} y^{3} z^{2})

= sec^{2} (5 + 2 x^{2} y^{3} z^{2}) \frac{\partial}{\partial x} (5 + 2 x^{2} y^{3} z^{2})

\frac{\partial}{\partial x} (5 + 2 x^{2} y^{3} z^{2}) = 4 y^{3} z^{2} x

= sec^{2} (5 + 2 x^{2} y^{3} z^{2}) \cdot 4 y^{3} z^{2} x

\frac{d}{d x} ((1 + sin (x)) \cdot cos (2 x))

\int \frac{2 x ^{2} + 3 x + 7}{x ^{3} + 3 x} d x

\frac{d y}{d x} = (cos (x)) e^{y + s i n (x)}, y (0) = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = - \frac{1}{5}, y^{^{'}} (0) = - \frac{8}{5}

3 e^{2 t} sin (4 t)

What is (\partial)/(\partial x)(tan(5+2x^2y^3z^2)) ?
The answer to (\partial)/(\partial x)(tan(5+2x^2y^3z^2)) is sec^2(5+2x^2y^3z^2)*4y^3z^2x